To determine the minimum of an equation, we derive the <span>equation using differential calculus twice (or simply </span><span>take the second derivative of the function). If the </span><span>second derivative is greater than 0, then it is minimum; </span><span>else, if it is less than 1, the function contains the </span><span>maximum. If the second derivative is zero, then the </span><span>inflection point </span><span>is</span><span> identified.</span>
Step-by-step explanation:
√a = 1√a so we can solve them easily:
b) 3√7 -√7= 3√7 - 1√7 =( 3-1)√7= 2√7
d) 5√6 - 2√6+√6= (5-2+1)√6 = 4√6
g) √2+2√2= 3√2
j) √5+5√5 - 3√5 = 3√5
k) 2√3 + √3 - 5√3= -2√3
I) 5√11 + 7√11 - √11 = 11√11
Notice the following pattern:
2 - 0 = 2
6 - 2 = 4
12 - 6 = 6
20 - 12 = 8
It's reasonable to assume that consecutive terms in the sequence differ by increasing multiples of 2, so that for the next number (call it x) we expect to see
x - 20 = 10 ==> x = 30
and for the number after that (call it y) we would see
y - x = y - 30 = 12 ==> y = 42
